- #1

MostlyHarmless

- 345

- 15

Let ##p## be a polynomial over a base field ##\mathbb{F}##. Let ##\mathbb{K}=\mathbb{F}(\alpha)## where ##\alpha \not\in\mathbb{F}## and ##\alpha## is a root of p. Now, suppose ##p## splits over ##\mathbb{K}##. Then ##\mathbb{K}## is the splitting field of ##p##. (?)

Is this argument sound? It seems like if it were, this is something that might have been mentioned in our discussion of splitting fields.

__Proof.__Let ##\mathbb{L}## be the splitting field of ##p## over ##\mathbb{F}##. Then ##\mathbb{L} \subseteq \mathbb{K}## since ##p## splits over ##\mathbb{K}##. Now, ##\alpha \in \mathbb{L}##, but by definition ##\mathbb{K} =\mathbb{F}(\alpha)## is the smallest field containing ##\mathbb{F}## and ##\alpha##. So ##\mathbb{K} \subseteq \mathbb{L}##. So mutual containment is established and ##\mathbb{L} = \mathbb{K}##. //Is this argument sound? It seems like if it were, this is something that might have been mentioned in our discussion of splitting fields.